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Math Help - [SOLVED] Function Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Function Question

    Question:
    The function f : x \mapsto x^2 + 5x + 8 is defined for the domain x \leq a, where a is a constant.
    (i) Express x^2 + 5x + 8 in the form (x + p)^2 + q.
    (ii) Find the smallest value of a for which f has an inverse.
    (iii) Find the domain of f^{-1} corresponding to this value of a.


    Attempt:
    (i) f(x) = x^2 + 5x + 8

    = (x + \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2 - 8

    = (x + \frac{5}{2})^2 + \frac{7}{4}

    p = \frac{5}{2} \mbox{  and  }q = \frac{7}{4}

    Is the domain = -\frac{5}{2} and range = \frac{7}{4} ?

    = (x + \underbrace{\frac{5}{2}}_{Domain})^2 + \underbrace{\frac{7}{4}}_{Range}
    Last edited by looi76; May 4th 2008 at 12:26 PM.
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  2. #2
    Kai
    Kai is offline
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    Draw the graph, easier to know domain and range,

    And remember, domain of f(x)= range of finverse(x)
    Range of f(x) = Domain of finverse(x)
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  3. #3
    MHF Contributor
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    Quote Originally Posted by looi76 View Post
    Question:
    The function f : x \mapsto x^2 + 5x + 8 is defined for the domain x \leq a, where a is a constant.
    (i) Express x^2 + 5x + 8 in the form (x + p)^2 + q.
    (ii) Find the smallest value of a for which f has an inverse.
    (iii) Find the domain of f^{-1} corresponding to this value of a.

    Attempt:
    (i) f(x) = x^2 + 5x + 8

    = (x + \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2 - 8

    = (x + \frac{5}{2})^2 + \frac{7}{4}

    p = \frac{5}{2} \mbox{ and }q = {7}{4}

    Is the domain = -\frac{5}{2} and range = \frac{7}{4} ?

    = (x + \underbrace{\frac{5}{2}}_{Domain})^2 + \underbrace{\frac{7}{4}}_{Range}
    For the function (x + \frac{5}{2})^2 + \frac{7}{4} the domain is x \leq -\frac{5}{2} and the range is y \geq \frac{7}{4}.
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